JABATAN KEJURUTERAAN ELEKTIUK MAKMAL TELEKOMUNIKASI ATIENUATOR AND IMPEDANCE TRANSFORMER 1.
Atteauator The purpose
of an Attenuator is to cause loss of a known dB.
An attenuatar is
a 2-port network consisting of resistors only. Here, you shall study
1.1
T-network Attenuatar
a) b)
Variable T-nctwork: Attenuataf
c)
Variable 1C-network
Attenuatar
T-ncnvork Attenultor
i
V,
I
a erlUa or The R.. Rb T-network is to be used
as
an attenuatar. The design equations may
be shown to be as follows: R
=N-IR. N+I
•
. . . . .............
2N
II.
(I)
N' _Ill. where Ro is the characteristic or image impedance of the T-network and N is =
the insertion loss ratio measured by V,IV,
.
Or in dB, take 2010g",N dB.
1.1.1
Experiment
a)
Calculation: The T-network is designed to be a 5
dB attenuatar.
If R" = 600 n, find R. and R•. b)
Measure R. and 14 as in
0)
To verify that the internal resistance of the Signal Generator II, is 600 n. Set signal generator to 10 kHz (say) and V, 6 volts (use oscilloscope). Co nnect a load resistor R,. (decade box) and Vary the resistance so that V,
the T-network. Compare with the calculation.
�
=
3
volu. What is the Rl. value?
1
d)
Attenuator Insertion Loss Measurement
T
R
i
V,
V,
I
I 1000 0 and V,
i)
For R,. = 100.200•
ii)
Calculate the attenuator insertion loss in dB or20logloV,IV1 dB You should get 5 dB for all cases of load resistance RI..
....•
600•
.. . .
=
6 volts. measure V, .
Can you explain why? 1.2
Vari.ble T-network Attenuator
a)
A 0 to 15 dB variable T-network Attenuator in stein of 5 dB is provided.
b)
Calculation: Fill up the Table below
,
dB
N
0
R.
R.
S 10 15
c)
The variable T-network attenuator is given below baving a SSDT
switch double throw) switch
, ,
a(OdB)
-1a (0 dB)
_____________
(5 dB)
b (5 dB) _-1.R.�1 c
(lOdBt"--�ri{=(10=}-.-::t_:..-C=J.I!",.-.-.4C dB)
d (IS
d)
(single
dB),-Ji�=d=}-"1f_-{==m:;_-
Calculate the values of
(IS
dB)
R.,. R... R...14,.R.z and R.z.
Measure aD resiston of the variable T-network Attenuator. Are they the same u calculated?
2
e)
Measure the insenion loss in dB and verifY th at it yields 0, 5, 10 and
1.3
Variable x-Network Attenuator.
-15 dB.
1
R,
R,
The R/, Rl x·nctwork is to be used as an attenuator. The design equations may be shown to be as follows: N +I · D Rt = N _ I '''tJ
IR. N'= 14 2N
a)
............. (2)
Calculation: Fill up the Table below: dB
N
R,
R,
I
2 4 8 .. ,
b)
Sketch the variable x-Network Attenuator having 0 to 1 5 dB in step of 1 dB, It consists of four 1t-networks connected i n cascade. Enter aU the resistor values. Show clearly the connection of the four SSDT switches.
c)
Measurement of the insertion loss in dB. Connect a 600 n load. Measure and verify the variable 1t-network Attenuator insertion loss in dB. namely O. 1.2.3• ......• 15 dB.
2
lmpedaace Traos(ormer
Consider the following circuit
600u
'9
11.= 1800
3
the aboyc c..15e maximum pawu transfer does not occur. The purpose of an impedance transfonner is to have impedance matching so that maximum power transfer can take place. In
a)
Calculation
i)
If Yo =6volts, R, = 600 n andR, If Yo 6 volts. Rg = 600 n andR{.
=
=
ii)
Show that the power transfer ratio
b)
An Impedance
R
=
600 n,
show that P_= 15 mW. 1800 n, show that Pr. = 11-25 mW.
Transfonner =
L
R, = 1800 !1
Z(f) R(f) + jX(f) =
i)
Derive an expression for ZifJ. hence its real partROO and imaginary part X(f) in terms of L, C and R,.
ill
Then. we can show that
"
c)
Calculation If the operating frequency!o is 30 kHz. calculate the values for L and C Measure the L and C values given in experiment (use an LCR meter).
d)
Measurement and Calculation of the Power Transfer RatiQ For frequencyf= 5, 10, 15, ......, 50 kHz and V, = 6 volts. Measure V, , which is a function of frequency, hence V,(I).
4
.
'
Clearly it is known that n "
- VI} _
P-.
R,
v.' ' :-
4R,
I(J)=.!L p
Experiment and 6lJ Up the Table below:
�;f
Pdf)
VIj) volts
t(f)
mW
5 10 15 20
"
25 30 35 40 45 50
e)
Plot t(f) against!
1.0 . 0.7S 0.6
. ...�.........;.........!. ·
.
.
.. ..
..
..
. . .
. . .
-}......... .........}.......................
...
· .. · .. · " . . . ... . . . .. ... ... . . . . .... . . . .. . .......... . ... . . . . . . .
,
,
,
,
,
.
'
. . .
. ....
..
,
.. .
.
. ..
....... . .
,
; ...... ! . ,..; .:: .! j : : : :j : . :j.........:: ., : . OZ·· - r ·,·· · · ·L ., L : : 10 15 ZO 25 30 35 4 .....
;
.. .
.
) .•.......
!
..•......
·
3.
!
,
.
. . . . . ............. . . .
. .
.
.
....••...
.
..•.•....
'
, .........•.........
.
.
.
.
I
.
.. . . . . . .
....•....
) .......••
..
[(kHz)
Some QuestioDS: i)
il) iii)
Why the Attenuator circuit consists of Resistors only? Design • variable T-network Attenuator having 0 to 50 dB in step of 10 dB and R. = 600 n. Why iIle Impedance Transformer consists ofL and e only?
Prof. Lu/031uly 1998
5
